Graphing Quadratics Notes and Worksheets - Lindsay Bowden - Free Printable

Let’s solve each equation step by step to find the zeros (roots) of each quadratic.

---

First Equation: Two Real Roots
Equation: \( x^2 + 2x - 3 = 0 \)

We can factor this:

Look for two numbers that multiply to -3 and add to +2 → those are +3 and -1.

So:
\( (x + 3)(x - 1) = 0 \)

Set each factor equal to zero:
→ \( x + 3 = 0 \) → \( x = -3 \)
→ \( x - 1 = 0 \) → \( x = 1 \)

✔ Zeros: -3 and 1

---

Second Equation: One Real Root
Equation: \( x^2 + 6x + 9 = 0 \)

This is a perfect square trinomial.

Factor:
\( (x + 3)^2 = 0 \)

Set equal to zero:
→ \( x + 3 = 0 \) → \( x = -3 \)

✔ Zero: -3 (only one, repeated root)

---

Third Equation: No Real Roots
Equation: \( (x - 1)^2 + 3 = 0 \)

Subtract 3 from both sides:
→ \( (x - 1)^2 = -3 \)

Now take square root of both sides:
→ \( x - 1 = \pm \sqrt{-3} \)

But square root of a negative number is not real — it’s imaginary.

So there are no real solutions.

✔ Zeros: None (in real numbers)

---

Final Answer:
For \( x^2 + 2x - 3 = 0 \): zeros are \( x = -3 \) and \( x = 1 \)
For \( x^2 + 6x + 9 = 0 \): zero is \( x = -3 \)
For \( (x - 1)^2 + 3 = 0 \): no real zeros